Bases of a Topology Examples 1

Recall from the Bases of a Topology page that if $(X, \tau)$ is a topological space then a base for the topology $\tau$ is a collection $\mathcal B \subseteq \tau$ such that every $U \in \tau$ can be written as a union of elements from $\mathcal B$, i.e., for all $U \in \tau$ we have that there exists a $\mathcal B^* \subseteq \mathcal B$ such that:

(1)

\begin{align} \quad U = \bigcup_{B \in \mathcal B^*} B \end{align}

We will now look at some more examples of bases for topologies.

Example 1

Consider the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the usual topology on $\mathbb{R}$. Show that $\mathcal B = \{ (a, b) : a, b \in \mathbb{R}, a < b \}$ is a base of $\tau$.

Notice that the open sets of $\mathbb{R}$ with respect to $\tau$ are the the empty set $\emptyset$ and whole set $\mathbb{R}$, open intervals, the unions of arbitrary collections of open intervals, and the intersections of finite collections of open intervals.

The empty set can be obtained from the base $\mathcal B$ by taking the empty union of elements from $\mathcal B$. Furthermore, the whole set, $\mathbb{R}$ can be obtained rather trivially as:

(2)

\begin{align} \quad \mathbb{R} = \bigcup_{a, b \in \mathbb{R}}_{a < b} (a, b) \end{align}

Any single open interval $(a, b) \in \mathbb{R}$ is clearly contained in $\mathcal B$ as the single union of $(a, b) \in \mathcal B$.

Now consider the union of an arbitrary collection of open intervals, $\{ U_i \}_{i \in I}$ where $U_i = (a, b)$ for some $a, b \in \mathbb{R}$, $a < b$ for each $i \in I$. Then the union $\bigcup_{i \in I} U_i$ is equal to the union of the intervals $U_i \in \mathcal B$.

Lastly, consider the intersection of a finite collection of open intervals. The intersection is either an open interval or the empty set, both of which can be obtained from taking unions of the open intervals in $\mathcal B$.

Example 2

Consider the set $X = \{a, b, c, d \}$ and the set $\mathcal B = \{ \{ a \}, \{c, d \}, \{a, b, c\} \}$. Determine whether there exists a topology $\tau$ on $X$ such that $\mathcal B$ is a base for $\tau$.

Note that in this example we are not implying that $\mathcal B$ is a base of $\tau$ since we don't even know if such a $\tau$ exists with $\mathcal B$ as a base of $\tau$.

All possible unions of elements from $\mathcal B$ are given below:

(3)

\begin{align} \quad \left \{ \bigcup_{B \in \mathcal B^*} : \mathcal B^* \subseteq \mathcal B \right \} = \{ \emptyset, \{ a \}, \{c, d \}, \{a, b, c \}, \{ a, c, d \}, X \} \end{align}

If $\tau$ is a topology generated by $\mathcal B$ then $\tau = \{ \emptyset, \{ a \}, \{c, d \}, \{a, b, c \}, \{ a, c, d \}, X \}$. Notice though that:

(4)

\begin{align} \quad \{c, d \} \cap \{a, b, c \} = \{ c \} \not \in \tau \end{align}

Therefore there exists no topology $\tau$ with $\mathcal B$ as a base.